A023537 -id:A023537 - cp's OEIS frontend

A027962 Duplicate of A023537.

1, 5, 13, 28, 54, 98, 171, 291, 487, 806, 1324, 2164, 3525, 5729, 9297, 15072

Offset: 2

dead

A104793 Triangle T(n,k) = A023537(n-k), n >= 1, 0 <= k < n, read by rows.

1, 5, 1, 13, 5, 1, 28, 13, 5, 1, 54, 28, 13, 5, 1, 98, 54, 28, 13, 5, 1, 171, 98, 54, 28, 13, 5, 1, 291, 171, 98, 54, 28, 13, 5, 1, 487, 291, 171, 98, 54, 28, 13, 5, 1, 806, 487, 291, 171, 98, 54, 28, 13, 5, 1, 1324, 806, 487, 291, 171, 98, 54, 28, 13, 5, 1

Offset: 1

Gary W. Adamson, Mar 26 2005

Repeatedly writing the sequence A023537 backwards.

			First few rows of the triangle are:
   1;
   5,  1;
  13,  5,  1;
  28, 13,  5,  1;
  54, 28, 13,  5, 1;
  98, 54, 28, 13, 5, 1; ...

G. C. Greubel, Rows n = 1..100 of triangle, flattened

Row sums are in A027963.

Cf. A104765.

Flat(List([1..12], n-> List([0..n-1], k-> Lucas(1, -1, n-k+4)[2] -3*n+3*k-7 ))); # G. C. Greubel, Jun 01 2019

[[Lucas(n-k+4) -(3*n-3*k+7): k in [0..n-1]]: n in [1..12]]; // G. C. Greubel, Jun 01 2019

Table[LucasL[n-k+4] -3*n+3*k-7, {n,1,12}, {k,0,n-1}]//Flatten (* G. C. Greubel, Jun 01 2019 *)

{T(n,k) = fibonacci(n-k+5) + fibonacci(n-k+3) -3*n +3*k - 7}; \\ G. C. Greubel, Jun 01 2019

[[lucas_number2(n-k+4, 1, -1) -3*n+3*k-7 for k in (0..n-1)] for n in (1..12)] # G. C. Greubel, Jun 01 2019

From Ralf Stephan, Apr 05 2009: (Start)
T(n,k) = Lucas(n-k+4) - (3*n - 3*k + 7).
T(n,k) = A023537(A004736(n, k+1)). (End)

Edited by Ralf Stephan, Apr 05 2009

A192744 Constant term of the reduction by x^2->x+1 of the polynomial p(n,x) defined below in Comments.

Original entry on oeis.org

1, 1, 3, 8, 29, 133, 762, 5215, 41257, 369032, 3676209, 40333241, 483094250, 6271446691, 87705811341, 1314473334832, 21017294666173, 357096406209005, 6424799978507178, 122024623087820183, 2439706330834135361, 51219771117454755544

Offset: 0

Clark Kimberling, Jul 09 2011

nonn

The titular polynomial is defined recursively by p(n,x)=x*p(n-1,x)+n! for n>0, where p(0,x)=1; see the Example.  For an introduction to polynomial reduction, see A192232.  The discussion at A192232 Comments continues here:
...
Let R(p,q,s) denote the "reduction of polynomial p by q->s" as defined at A192232.  Suppose that q(x)=x^k for some k>0 and that s(x)=s(k,0)*x^(k-1)+s(k,1)*x^(k-2)+...+s(k,k-2)x+s(k,k-1).
...
First, we shall take p(x)=x^n, where n>=0; the results will be used to formulate R(p,q,s) for general p.  Represent R(x^n,q,s) by
...
R(x^n)=s(n,0)*x^(k-1)+s(n,1)*x^(k-2)+...+s(n,k-2)*x+s(n,k-1).
...
Then each of the sequences u(n)=s(n,h), for h=0,1,...,k-1, satisfies this linear recurrence relation:
...
u(n)=s(k,0)*u(n-1)+s(k,1)*u(n-2)+...+s(k,k-2)*u(n-k-1)+s(k,k-1)*u(n-k), with initial values tabulated here:
...
n: ..s(n,0)...s(n,1)..s(n,2).......s(n,k-2)..s(n,k-1)
0: ....0........0.......0..............0.......1
1: ....0........0.......0..............1.......0
...
k-2: ..0........1.......0..............0.......0
k-1: ..0........0.......0..............0.......0
k: ..s(k,0)...s(k,1)..s(k,2).......s(k,k-2)..s(k,k-1)
...
That completes the formulation for p(x)=x^n.  Turning to the general case, suppose that
...
p(n,x)=p(n,0)*x^n+p(n,1)*x^(n-1)+...+p(n,n-1)*x+p(n,n)
...
is a polynomial of degree n>=0.  Then the reduction denoted by (R(p(n,x) by x^k -> s(x)) is the polynomial of degree k-1 given by the matrix product P*S*X, where P=(p(n,0)...p(n,1).........p(n-k)...p(n,n-k+1); X has all 0's except for main diagonal (x^(k-1), x^(k-2)...x,1); and S has
...
row 1: ... s(n,0) ... s(n,1) ...... s(n,k-2) . s(n,k-1)
row 2: ... s(n-1,0) . s(n-1,1) .... s(n-1,k-2) s(n-1,k-1)
...
row n-k+1: s(k,0).... s(k,1) ...... s(k,k-2) ..s(k,k-1)
row n-k+2: p(n,n-k+1) p(n,n-k+2) .. p(n,n-1) ..p(n,n)
*****
As a class of examples, suppose that (v(n)), for n>=0, is a sequence, that p(0,x)=1, and p(n,x)=v(n)+p(n-1,x) for n>0.  If q(x)=x^2 and s(x)=x+1, and we write the reduction R(p(n,x)) as u1(n)*x+u2(n), then the sequences u1 and u2 are convolutions with the Fibonacci sequence, viz., let F=(0,1,1,2,3,5,8,...)=A000045 and let G=(1,0,1,1,2,3,5,8...); then u1=G**v and u2=F**v, where ** denotes convolution.  Examples (with a few exceptions for initial terms):
...
If v(n)=n! then u1=A192744, u2=A192745.
If v(n)=n+1 then u1=A000071, u2=A001924.
If v(n)=2n then u1=A014739, u2=A027181.
If v(n)=2n+1 then u1=A001911, u2=A001891.
If v(n)=3n+1 then u1=A027961, u2=A023537.
If v(n)=3n+2 then u1=A192746, u2=A192747.
If v(n)=3n then u1=A154691, u2=A192748.
If v(n)=4n+1 then u1=A053311, u2=A192749.
If v(n)=4n+2 then u1=A192750, u2=A192751.
If v(n)=4n+3 then u1=A192752, u2=A192753.
If v(n)=4n then u1=A147728, u2=A023654.
If v(n)=5n+1 then u1=A192754, u2=A192755.
If v(n)=5n then u1=A166863, u2=A192756.
If v(n)=floor((n+1)tau) then u1=A192457, u2=A023611.
If v(n)=floor((n+2)/2) then u1=A052952, u2=A129696.
If v(n)=floor((n+3)/3) then u1=A004695, u2=A178982.
If v(n)=floor((n+4)/4) then u1=A080239, u2=A192758.
If v(n)=floor((n+5)/5) then u1=A124502, u2=A192759.
If v(n)=n+2 then u1=A001594, u2=A192760.
If v(n)=n+3 then u1=A022318, u2=A192761.
If v(n)=n+4 then u1=A022319, u2=A192762.
If v(n)=2^n then u1=A027934, u2=A008766.
If v(n)=3^n then u1=A106517, u2=A094688.

			The first five polynomials and their reductions:
1 -> 1
1+x -> 1+x
2+x+x^2 -> 3+2x
6+2x+x^2+x^3 -> 8+5x
24+6x+2x^2+x^3+x^4 -> 29+13x, so that
A192744=(1,1,3,8,29,...) and A192745=(0,1,2,5,13,...).

Cf. A192232.

A192744p := proc(n,x)
    option remember;
    if n = 0 then
        1;
    else
        x*procname(n-1,x)+n! ;
        expand(%) ;
    end if;
end proc:
A192744 := proc(n)
    local p;
    p := A192744p(n,x) ;
    while degree(p,x) > 1 do
        p := algsubs(x^2=x+1,p) ;
        p := expand(p) ;
    end do:
    coeftayl(p,x=0,0) ;
end proc: # R. J. Mathar, Dec 16 2015

q = x^2; s = x + 1; z = 40;
p[0, n_] := 1; p[n_, x_] := x*p[n - 1, x] + n!;
Table[Expand[p[n, x]], {n, 0, 7}]
reduce[{p1_, q_, s_, x_}] :=
FixedPoint[(s PolynomialQuotient @@ #1 +
       PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]
t = Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];
u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}]
  (* A192744 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}]
  (* A192745 *)

G.f.: (1-x)/(1-x-x^2)/Q(0), where Q(k)= 1 - x*(k+1)/(1 - x*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, May 20 2013
Conjecture: a(n) +(-n-2)*a(n-1) +2*(n-1)*a(n-2) +3*a(n-3) +(-n+2)*a(n-4)=0. - R. J. Mathar, May 04 2014
Conjecture: (-n+2)*a(n) +(n^2-n-1)*a(n-1) +(-n^2+3*n-3)*a(n-2) -(n-1)^2*a(n-3)
=0. - R. J. Mathar, Dec 16 2015

A101220 a(n) = Sum_{k=0..n} Fibonacci(n-k)*n^k.

Original entry on oeis.org

0, 1, 3, 14, 91, 820, 9650, 140601, 2440317, 49109632, 1123595495, 28792920872, 816742025772, 25402428294801, 859492240650847, 31427791175659690, 1234928473553777403, 51893300561135516404, 2322083099525697299278

Offset: 0

Ross La Haye, Dec 14 2004

In what follows a(i,j,k) denotes a three-dimensional array, the terms a(n) are defined as a(n,n,n) in that array. - Joerg Arndt, Jan 03 2021
Previous name was: Three-dimensional array: a(i,j,k) = expansion of x*(1 + (i-j)*x)/((1-j*x)*(1-x-x^2)), read by a(n,n,n).
a(i,j,k) = the k-th value of the convolution of the Fibonacci numbers (A000045) with the powers of i = Sum_{m=0..k} a(i-1,j,m), both for i = j and i > 0; a(i,j,k) = a(i-1,j,k) + a(j,j,k-1), for i,k > 0; a(i,1,k) = Sum_{m=0..k} a(i-1,0,m), for i > 0. With F = Fibonacci and L = Lucas, then a(1,1,k) = F(k+2) - 1; a(2,1,k) = F(k+3) - 2; a(3,1,k) = L(k+2) - 3; a(4,1,k) = 4*F(k+1) + F(k) - 4; a(1,2,k) = 2^k - F(k+1); a(2,2,k) = 2^(k+1) - F(k+3); a(3,2,k) = 3(2^k - F(k+2)) + F(k); a(4,2,k) = 2^(k+2) - F(k+4) - F(k+2); a(1,3,k) = (3^k + L(k-1))/5, for k > 0; a(2,3,k) = (2 * 3^k - L(k)) /5, for k > 0; a(3,3,k) = (3^(k+1) - L(k+2))/5; a(4,3,k) = (4 * 3^k - L(k+2) - L(k+1))/5, etc..

			a(1,3,3) = 6 because a(1,3,0) = 0, a(1,3,1) = 1, a(1,3,2) = 2 and 4*2 - 2*1 - 3*0 = 6.

G. C. Greubel, Table of n, a(n) for n = 0..385
Eric Weisstein's World of Mathematics, Fibonacci Number
Eric Weisstein's World of Mathematics, Lucas Number

a(0, j, k) = A000045(k).
a(1, 2, k+1) - a(1, 2, k) = A099036(k).
a(3, 2, k+1) - a(3, 2, k) = A104004(k).
a(4, 2, k+1) - a(4, 2, k) = A027973(k).
a(1, 3, k+1) - a(1, 3, k) = A099159(k).
a(i, 0, k) = A109754(i, k).
a(i, i+1, 3) = A002522(i+1).
a(i, i+1, 4) = A071568(i+1).
a(2^i-2, 0, k+1) = A118654(i, k), for i > 0.
Sequences of the form a(n, 0, k): A000045(k+1) (n=1), A000032(k) (n=2), A000285(k-1) (n=3), A022095(k-1) (n=4), A022096(k-1) (n=5), A022097(k-1) (n=6), A022098(k-1) (n=7), A022099(k-1) (n=8), A022100(k-1) (n=9), A022101(k-1) (n=10), A022102(k-1) (n=11), A022103(k-1) (n=12), A022104(k-1) (n=13), A022105(k-1) (n=14), A022106(k-1) (n=15), A022107(k-1) (n=16), A022108(k-1) (n=17), A022109(k-1) (n=18), A022110(k-1) (n=19), A088209(k-2) (n=k-2), A007502(k) (n=k-1), A094588(k) (n=k).
Sequences of the form a(1, n, k): A000071(k+2) (n=1), A027934(k-1) (n=2), A098703(k) (n=3).
Sequences of the form a(2, n, k): A001911(k) (n=1), A008466(k+1) (n=2), A106517(k-1) (n=3).
Sequences of the form a(3, n, k): A027961(k) (n=1), A094688(k) (n=3).
Sequences of the form a(4, n, k): A053311(k-1) (n=1), A027974(k-1) (n=2).
Cf. A001622, A001891, A001924, A023537, A104161, A106517.

A101220:= func< n | (&+[n^k*Fibonacci(n-k): k in [0..n]]) >;
[A101220(n): n in [0..30]]; // G. C. Greubel, Jun 01 2025

Join[{0}, Table[Sum[Fibonacci[n-k]*n^k, {k, 0, n}], {n, 1, 20}]] (* Vaclav Kotesovec, Jan 03 2021 *)

a(n)=sum(k=0,n,fibonacci(n-k)*n^k) \\ Joerg Arndt, Jan 03 2021

def A101220(n): return sum(n^k*fibonacci(n-k) for k in range(n+1))
print([A101220(n) for n in range(31)]) # G. C. Greubel, Jun 01 2025

a(i, j, 0) = 0, a(i, j, 1) = 1, a(i, j, 2) = i+1; a(i, j, k) = ((j+1)*a(i, j, k-1)) - ((j-1)*a(i, j, k-2)) - (j*a(i, j, k-3)), for k > 2.
a(i, j, k) = Fibonacci(k) + i*a(j, j, k-1), for i, k > 0.
a(i, j, k) = (Phi^k - (-Phi)^-k + i(((j^k - Phi^k) / (j - Phi)) - ((j^k - (-Phi)^-k) / (j - (-Phi)^-1)))) / sqrt(5), where Phi denotes the golden mean/ratio (A001622).
i^k = a(i-1, i, k) + a(i-2, i, k+1).
A104161(k) = Sum_{m=0..k} a(k-m, 0, m).
a(i, j, 0) = 0, a(i, j, 1) = 1, a(i, j, 2) = i+1, a(i, j, 3) = i*(j+1) + 2; a(i, j, k) = (j+2)*a(i, j, k-1) - 2*j*a(i, j, k-2) - a(i, j, k-3) + j*a(i, j, k-4), for k > 3. a(i, j, 0) = 0, a(i, j, 1) = 1; a(i, j, k) = a(i, j, k-1) + a(i, j, k-2) + i * j^(k-2), for k > 1.
G.f.: x*(1 + (i-j)*x)/((1-j*x)*(1-x-x^2)).
a(n, n, n) = Sum_{k=0..n} Fibonacci(n-k) * n^k. - Ross La Haye, Jan 14 2006
Sum_{m=0..k} binomial(k,m)*(i-1)^m = a(i-1,i,k) + a(i-2,i,k+1), for i > 1. - Ross La Haye, May 29 2006
From Ross La Haye, Jun 03 2006: (Start)
a(3, 3, k+1) - a(3, 3, k) = A106517(k).
a(1, 1, k) = A001924(k) - A001924(k-1), for k > 0.
a(2, 1, k) = A001891(k) - A001891(k-1), for k > 0.
a(3, 1, k) = A023537(k) - A023537(k-1), for k > 0.
Sum_{j=0..i+1} a(i-j+1, 0, j) - Sum_{j=0..i} a(i-j, 0, j) = A001595(i). (End)
a(i,j,k) = a(j,j,k) + (i-j)*a(j,j,k-1), for k > 0.
a(n) ~ n^(n-1). - Vaclav Kotesovec, Jan 03 2021

New name from Joerg Arndt, Jan 03 2021

A027960 'Lucas array': triangular array T read by rows.

Original entry on oeis.org

1, 1, 3, 1, 1, 3, 4, 4, 1, 1, 3, 4, 7, 8, 5, 1, 1, 3, 4, 7, 11, 15, 13, 6, 1, 1, 3, 4, 7, 11, 18, 26, 28, 19, 7, 1, 1, 3, 4, 7, 11, 18, 29, 44, 54, 47, 26, 8, 1, 1, 3, 4, 7, 11, 18, 29, 47, 73, 98, 101, 73, 34, 9, 1, 1, 3, 4, 7, 11, 18, 29, 47, 76, 120, 171, 199, 174, 107, 43, 10, 1

Offset: 0

Clark Kimberling

The k-th row contains 2k+1 numbers.
Columns in the right half consist of convolutions of the Lucas numbers with the natural numbers.
T(n,k) = number of strings s(0),...,s(n) such that s(n)=n-k. s(0) in {0,1,2}, s(1)=1 if s(0) in {1,2}, s(1) in {0,1,2} if s(0)=0 and for 1 <= i <= n, s(i) = s(i-1)+d, with d in {0,2} if s(i)=2i, in {0,1,2} if s(i)=2i-1, in {0,1} if 0 <= s(i) <= 2i-2.

			                           1
                       1,  3,  1
                   1,  3,  4,  4,  1
               1,  3,  4,  7,  8,  5,   1
           1,  3,  4,  7, 11, 15, 13,   6,  1
        1, 3,  4,  7, 11, 18, 26, 28,  19,  7,  1
     1, 3, 4,  7, 11, 18, 29, 44, 54,  47, 26,  8, 1
  1, 3, 4, 7, 11, 18, 29, 47, 73, 98, 101, 73, 34, 9, 1

Nathaniel Johnston, Table of n, a(n) for n = 0..10000

Central column is the Lucas numbers without initial 2: A000204.
Columns in the right half include A027961, A027962, A027963, A027964, A053298.
Bisection triangles are in A026998 and A027011.
Row sums: A036563, A153881 (alternating sign).
Sums include: A023537, A027973, A027974, A027975, A027976, A027978, A027979, A027980, A027981, A027982, A027984, A027985, A027986, A027987, A053298.
Diagonals of the form T(n, 2*n-p): A000012 (p=0), A000027 (p=1), A034856 (p=2), A027965 (p=3), A027966 (p=4), A027967 (p=5), A027968 (p=6), A027969 (p=7), A027970 (p=8), A027971 (p=9), A027972 (p=10).
Diagonals of the form T(n, n+p): A000032 (p=0), A027961 (p=1), A023537 (p=2), A027963 (p=3), A027964 (p=4), A053298 (p=5), A027002 U A027018 (p=6), A027007 U A027014 (p=7), A027003 U A027019 (p=8).
Cf. A000032, A004396, A026998, A027011, A027977, A075193, A168616.

function T(n,k) // T = A027960
      if k le n then return Lucas(k+1);
      elif k gt 2*n then return 0;
      else return T(n-1, k-2) + T(n-1, k-1);
      end if;
end function;
[T(n,k): k in [0..2*n], n in [0..12]]; // G. C. Greubel, Jun 08 2025

T:=proc(n,k)option remember:if(k=0 or k=2*n)then return 1:elif(k=1)then return 3:else return T(n-1,k-2) + T(n-1,k-1):fi:end:
for n from 0 to 6 do for k from 0 to 2*n do print(T(n,k));od:od: # Nathaniel Johnston, Apr 18 2011

(* First program *)
t[, 0] = 1; t[, 1] = 3; t[n_, k_] /; (k == 2*n) = 1; t[n_, k_] := t[n, k] = t[n-1, k-2] + t[n-1, k-1]; Table[t[n, k], {n, 0, 8}, {k, 0, 2*n}] // Flatten (* Jean-François Alcover, Dec 27 2013 *)
(* Second program *)
f[n_, k_]:= f[n,k]= Sum[Binomial[2*n-k+j,j]*LucasL[2*(k-n-j)], {j,0,k-n-1}];
A027960[n_, k_]:= LucasL[k+1] - f[n,k]*Boole[k>n];
Table[A027960[n,k], {n,0,12}, {k,0,2*n}]//Flatten (* G. C. Greubel, Jun 08 2025 *)

T(r,n)=if(r<0||n>2*r,return(0)); if(n==0||n==2*r,return(1)); if(n==1,3,T(r-1,n-1)+T(r-1,n-2)) /* Ralf Stephan, May 04 2005 */

@CachedFunction
def T(n, k): # T = A027960
    if (k>2*n): return 0
    elif (kG. C. Greubel, Jun 01 2019; Jun 08 2025

T(n, k) = Lucas(k+1) for k <= n, otherwise the (2n-k)th coefficient of the power series for (1+2*x)/{(1-x-x^2)*(1-x)^(k-n)}.
Recurrence: T(n, 0)=T(n, 2n)=1 for n >= 0; T(n, 1)=3 for n >= 1; and for n >= 2, T(n, k) = T(n-1, k-2) + T(n-1, k-1) for 2 <= k <= 2*n-1.
From G. C. Greubel, Jun 08 2025: (Start)
T(n, k) = A000032(k+1) - f(n, k)*[k > n], where f(n, k) = Sum_{j=0..k-n-1} binomial(2*n -k+j, j)*A000032(2*(k-n-j)).
Sum_{k=0..A004396(n+1)} T(n-k, k) = A027975(n).
Sum_{k=0..n} T(n, k) = A027961(n).
Sum_{k=0..2*n} T(n, k) = A168616(n+2) + 2.
Sum_{k=n+1..2*n} (-1)^k*T(n, k) = A075193(n-1), n >= 1. (End)

Edited by Ralf Stephan, May 04 2005

A027961 a(n) = Lucas(n+2) - 3.

Original entry on oeis.org

1, 4, 8, 15, 26, 44, 73, 120, 196, 319, 518, 840, 1361, 2204, 3568, 5775, 9346, 15124, 24473, 39600, 64076, 103679, 167758, 271440, 439201, 710644, 1149848, 1860495, 3010346, 4870844, 7881193, 12752040, 20633236, 33385279, 54018518

Offset: 1

Clark Kimberling

Sum of the first n Lucas numbers, that is, A000204(1) to A000204(n). - T. D. Noe, Oct 10 2005

G. C. Greubel, Table of n, a(n) for n = 1..2500
E. S. Egge and T. Mansour, Restricted permutations, Fibonacci numbers and k-generalized Fibonacci numbers, arXiv:math/0203226 [math.CO], 2002.
Index entries for linear recurrences with constant coefficients, signature (2,0,-1).

T(n, n+1), T given by A027960.

Cf. A000045, A000204, A023537, A101220.

List([1..40], n-> Lucas(1, -1, n+2)[2] -3 ); # G. C. Greubel, Jun 01 2019

[Lucas(n+2)-3: n in [1..40]]; // Vincenzo Librandi, Apr 16 2011

a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]+a[n-2]+3 od: seq(a[n],n=1..40); # Miklos Kristof, Mar 09 2005
with(combinat): seq(fibonacci(n)+fibonacci(n+2)-3, n=2..40); # Zerinvary Lajos, Jan 31 2008
g:=(1+z^2)/(1-z-z^2): gser:=series(g, z=0, 43): seq(coeff(gser, z, n)-3, n=3..40); # Zerinvary Lajos, Jan 09 2009

LucasL[Range[3, 40]] - 3 (* Alonso del Arte, Sep 26 2013 *)

vector(40, n, fibonacci(n+3) +fibonacci(n+1) -3) \\ G. C. Greubel, Dec 18 2017

first(n) = Vec(x*(1+2*x)/((1-x)*(1-x-x^2)) + O(x^(n+1))) \\ Iain Fox, Dec 19 2017

[lucas_number2(n+2, 1, -1) -3 for n in (1..40)] # G. C. Greubel, Jun 01 2019

a(0) = 0, a(1) = 1, a(n) = a(n-1) + a(n-2) + 3.
a(n) = A000204(n+2) - 3 = A000045(2n+4)/A000045(n+2) - 3. - Benoit Cloitre, Jan 05 2003
G.f.: x*(1+2*x)/((1-x)*(1-x-x^2)). Differences of A023537. - Ralf Stephan, Feb 07 2004
a(n) = A101220(3, 1, n). - Ross La Haye, Jan 28 2005
a(n) = F(n) + F(n+2) - 3, n >= 2, where F(n) is the n-th Fibonacci number. - Zerinvary Lajos, Jan 31 2008
a(n) = Sum_{k=1..n} ((-1/phi)^k + (phi)^k) where phi = 1/2+1/2*sqrt(5). - Dimitri Papadopoulos, Jan 07 2016
a(n) = 2*a(n-1)-a(n-3) for n>3. - Wesley Ivan Hurt, Jan 07 2016

A027974 a(n) = Sum_{k=1..n+1} A027960(n+1, n+1+k).

Original entry on oeis.org

1, 5, 14, 35, 81, 180, 389, 825, 1726, 3575, 7349, 15020, 30561, 61965, 125294, 252795, 509161, 1024100, 2057549, 4130225, 8284926, 16609455, 33282989, 66669660, 133507081, 267285605, 535010414, 1070731475, 2142612801, 4287086100, 8577182549, 17159235945

Offset: 0

Clark Kimberling

The former name, a(n) = Sum_{i=0..n} Sum_{j=0..i} T(i,j), T given by A027960, was in error with the data given. [This double summation gives A023537(n+1), or A027960(n+2, n+4) for n >= 0]. - G. C. Greubel, Jun 08 2025

G. C. Greubel, Table of n, a(n) for n = 0..1000
Philipp Emanuel Weidmann, The Sequencer OEIS Survey
Susanne Wienand, Suggestion for a proof of Philipp Emanuel Weidman's conjecture concerning A027983
Index entries for linear recurrences with constant coefficients, signature (3,-1,-2).

Cf. A000032, A000045, A027960, A101220.

List([0..30], n-> 2^(n+3) - Lucas(1,-1,n+4)[2]); # G. C. Greubel, Sep 26 2019

[2^(n+3) - Lucas(n+4): n in [0..30]]; // G. C. Greubel, Sep 26 2019

with(combinat); f:=fibonacci; seq(2^(n+3) - f(n+5) - f(n+3), n=0..30); # G. C. Greubel, Sep 26 2019

Table[2^(n+3) - LucasL[n+4], {n,0,30}] (* G. C. Greubel, Sep 26 2019 *)

vector(31, n, f=fibonacci; 2^(n+2) - f(n+4) - f(n+2)) \\ G. C. Greubel, Sep 26 2019

def A027974(n): return 2**(n+3) - lucas_number2(n+4,1,-1)
[A027974(n) for n in range(31)] # G. C. Greubel, Sep 26 2019; Jun 08 2025

a(n) = 2^(n+3) - Fibonacci(n+5) - Fibonacci(n+3).
a(n) = A101220(4, 2, n+1).
G.f.: (1+2*x)/((1-2*x)*(1-x-x^2)). - R. J. Mathar, Sep 22 2008
a(n) = 2*a(n-1) + A000032(n+1). - David A. Corneth, Apr 16 2015
a(n) = 3*a(n-1) - a(n-2) - 2*a(n-3). - Colin Barker, Feb 17 2016
From G. C. Greubel, Jun 08 2025: (Start)
a(n) = 2^(n+3) - A000032(n+4).
E.g.f.: 8*exp(2*x) - exp(x/2)*( 7*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2) ). (End)

A137176 Hyperlucas number array T(r,n) = L(n)^(r), read by ascending antidiagonals (r >= 0, n >= 0).

Original entry on oeis.org

0, 0, 1, 0, 1, 3, 0, 1, 4, 4, 0, 1, 5, 8, 7, 0, 1, 6, 13, 15, 11, 0, 1, 7, 19, 28, 26, 18, 0, 1, 8, 26, 47, 54, 44, 29, 0, 1, 9, 34, 73, 101, 98, 73, 47, 0, 1, 10, 43, 107, 174, 199, 171, 120, 76, 0, 1, 11, 53, 150, 281, 373, 370, 291, 196, 123

Offset: 0

Jonathan Vos Post, Apr 04 2008

In Theorem 17, Dil and Mezo (2008) connect the hyperlucas numbers (this array) with the incomplete Lucas numbers (A324242). - Petros Hadjicostas, Sep 03 2019

			The array T(r,n) = L(n)^(r) begins:
.....|n=0|n=1|.n=2|.n=3|.n=4.|.n=5.|..n=6.|.n=7..|..n=8..|..n=9..|.n=10..|.in.OEIS
r=0..|.0.|.1.|..3.|..4.|...7.|..11.|...18.|...29.|....47.|....76.|...123.|.A000204
r=1..|.0.|.1.|..4.|..8.|..15.|..26.|...44.|...73.|...120.|...196.|...319.|.A027961
r=2..|.0.|.1.|..5.|.13.|..28.|..54.|...98.|..171.|...291.|...487.|...806.|.A023537
r=3..|.0.|.1.|..6.|.19.|..47.|.101.|..199.|..370.|...661.|..1148.|..1954.|.A027963
r=4..|.0.|.1.|..7.|.26.|..73.|.174.|..373.|..743.|..1404.|..2552.|..4506.|.A027964
r=5..|.0.|.1.|..8.|.34.|.107.|.281.|..654.|.1397.|..2801.|..5353.|..9859.|.A053298
r=6..|.0.|.1.|..9.|.43.|.150.|.431.|.1085.|.2482.|..5283.|.10636.|.20495.|.new
r=7..|.0.|.1.|.10.|.53.|.203.|.634.|.1719.|.4201.|..9484.|.20120.|.40615.|.new
r=8..|.0.|.1.|.11.|.64.|.267.|.901.|.2620.|.6821.|.16305.|.36425.|.77040.|.new
r=9..|.0.|.1.|.12.|.76.|.343.|1244.|.3864.|10685.|.26990.|.63415.|140455.|.new
For example, T(4,5) = L(5)^(4) = L(0)^(3) + L(1)^(3) + L(2)^(3) + L(3)^(3) + L(4)^(3) + L(5)^(3) = 0 + 1 + 6 + 19 + 47 + 101 = 174. - _Petros Hadjicostas_, Sep 03 2019

Ayhan Dil and Istvan Mezo, A Symmetric Algorithm for Hyperharmonic and Fibonacci Numbers, arXiv:0803.4388 [math.NT], 2008.
Ayhan Dil and Istvan Mezo, A Symmetric Algorithm for Hyperharmonic and Fibonacci Numbers, Applied Mathematics and Computation 206(2) (2008), 942-951.
Eric W. Weisstein, Steenrod Algebra.

Cf. A000204, A027961, A023537, A027963, A027964, A053298, A123736, A136431.

Cf. A038730, A038792, and A134511 for incomplete Fibonacci sequences, and A324242 for incomplete Lucas sequences.

L:= proc(r, n) option remember; `if`(n=0, 0, `if`(r=0,
      `if`(n<3, 2*n-1, L(0, n-2)+L(0, n-1)), L(r-1, n)+L(r, n-1)))
    end:
seq(seq(L(d-n, n), n=0..d), d=0..12);  # Alois P. Heinz, Sep 03 2019

L[r_, n_] := L[r, n] = If[n == 0, 0, If[r == 0, If[n < 3, 2n-1, L[0, n-2] + L[0, n-1]], L[r-1, n] + L[r, n-1]]];
Table[L[d-n, n], {d, 0, 12}, {n, 0, d}] // Flatten (* Jean-François Alcover, Sep 26 2019, from Maple *)

T(r,n) = L(n)^(r) = Apply partial sum operator r times to Lucas numbers A000204.
From Petros Hadjicostas, Sep 03 2019: (Start)
T(r, n) = L(n)^(r) = Sum_{k = 0..n} L(k)^(r-1) for r >= 1, with T(0,n) = L(n)^(0) = L(n) = A000204(n), T(r,0) = L(0)^(r) = 0, and T(r,1) = L(1)^(r) = 1. (See Definition 13 in Dil and Mezo (2008).)
G.f. for row r: Sum_{n >= 0} L(n)^(r)*t^n = t * (1+2*t)/((1-t-t^2) * (1-t)^r). (Corrected from Proposition 14 in Dil and Mezo (2008).)
(End)

A033818 Convolution of natural numbers n >= 1 with Lucas numbers L(k) for k >= -2.

Original entry on oeis.org

3, 5, 9, 14, 22, 34, 53, 83, 131, 208, 332, 532, 855, 1377, 2221, 3586, 5794, 9366, 15145, 24495, 39623, 64100, 103704, 167784, 271467, 439229, 710673, 1149878, 1860526, 3010378, 4870877, 7881227, 12752075, 20633272, 33385316, 54018556, 87403839, 141422361, 228826165, 370248490

Offset: 1

Wolfdieter Lang

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,-2,-1,1).

Cf. A000032, A000045, A023537, A023548, A033811.

List([1..50], n-> Lucas(1,-1,n+1)[2] +n-1) # G. C. Greubel, Jun 01 2019

[Lucas(n+1) +n-1: n in [1..50]]; // G. C. Greubel, Jun 01 2019

LinearRecurrence[{3,-2,-1,1}, {3,5,9,14}, 50] (* Vladimir Joseph Stephan Orlovsky, Jan 28 2011, modified by G. C. Greubel, Jun 01 2019 *)
Table[LucasL[n+1] +n-1, {n,1,50}] (* G. C. Greubel, Jun 01 2019 *)

{a(n) = fibonacci(n+2) + fibonacci(n) + n-1}; \\ G. C. Greubel, Jun 01 2019

[lucas_number2(n+1,1,-1) +n-1 for n in (1..50)] # G. C. Greubel, Jun 01 2019

a(n) = L(1)*(F(n+1) - 1) + L(0)*F(n) - L(-1)*n, F(n): Fibonacci (A000045), L(n): Lucas (A000032) with L(-n) = (-1)^n*L(n).
G.f.: x*(3-4*x)/((1-x-x^2)*(1-x)^2).
a(n) = Lucas(n+1) + n - 1. - G. C. Greubel, Jun 01 2019

Terms a(31) onward added by G. C. Greubel, Jun 01 2019

A023550 Convolution of natural numbers >= 2 and (F(2), F(3), F(4), ...).

Original entry on oeis.org

2, 7, 16, 32, 59, 104, 178, 299, 496, 816, 1335, 2176, 3538, 5743, 9312, 15088, 24435, 39560, 64034, 103635, 167712, 271392, 439151, 710592, 1149794, 1860439, 3010288, 4870784, 7881131, 12751976, 20633170, 33385211, 54018448, 87403728, 141422247, 228826048

Offset: 1

Clark Kimberling

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3, -2, -1, 1).

Cf. A000032, A023537.

List([1..40], n-> Lucas(1, -1, n+4)[2] -2*n-7 ); # G. C. Greubel, Jun 01 2019

[Lucas(n+4) - 2*n - 7 : n in [1..40]]; // G. C. Greubel, Jun 01 2019

LinearRecurrence[{3,-2,-1,1}, {2,7,16,32}, 40] (* Vladimir Joseph Stephan Orlovsky, Feb 11 2012 *)

Vec(x*(2-x)*(1+x)/((1-x)^2*(1-x-x^2)) + O(x^40)) \\ Colin Barker, Mar 11 2017

vector(40, n, fibonacci(n+5) + fibonacci(n+3) -2*n-7) \\ G. C. Greubel, Jun 01 2019

[lucas_number2(n+4,1,-1) -2*n-7 for n in (1..40)] # G. C. Greubel, Jun 01 2019

a(n) = A023537(n) + 2*n.
From Colin Barker, Mar 11 2017: (Start)
G.f.: x*(2-x)*(1+x) / ((1-x)^2*(1-x-x^2)).
a(n) = (2^(-1-n)*((1-sqrt(5))^n*(-15+7*sqrt(5)) + (1+sqrt(5))^n*(15+7*sqrt(5)))) / sqrt(5) - 2*n - 7.
a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4) for n>4.
(End)
a(n) = Lucas(n+4) - 2*n - 7. - G. C. Greubel, Jun 01 2019

cp's OEIS Frontend

A027962 Duplicate of A023537.

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A104793 Triangle T(n,k) = A023537(n-k), n >= 1, 0 <= k < n, read by rows.

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A192744 Constant term of the reduction by x^2->x+1 of the polynomial p(n,x) defined below in Comments.

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A101220 a(n) = Sum_{k=0..n} Fibonacci(n-k)*n^k.

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A027960 'Lucas array': triangular array T read by rows.

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A027961 a(n) = Lucas(n+2) - 3.

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A027974 a(n) = Sum_{k=1..n+1} A027960(n+1, n+1+k).